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fix: 给kook适配器实现send_streaming方法(降级到普通文本消息发送)来修复启用了流逝输出后可能llm消息发不出去的bug #7749

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fix: 给kook适配器实现send_streaming方法(降级到普通文本消息发送)来修复启用了流逝输出后可能llm消息发不出去的bug #7749

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18 changes: 17 additions & 1 deletion astrbot/core/platform/sources/kook/kook_event.py
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Original file line number Diff line number Diff line change
@@ -1,6 +1,6 @@
import asyncio
import json
from collections.abc import Coroutine
from collections.abc import AsyncGenerator, Coroutine
from pathlib import Path
from typing import Any

Expand Down Expand Up @@ -208,3 +208,19 @@ async def send(self, message: MessageChain):
logger.error(f"[kook] {err_msg}")

await super().send(message)

async def send_streaming(
self, generator: AsyncGenerator[MessageChain, None], use_fallback: bool = False
):
"""KOOK 平台不支持流式输出, 调用此方法将回退到普通`send`方法"""
buffer = None
async for chain in generator:
if not buffer:
buffer = chain
Comment on lines +218 to +219
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@sourcery-ai sourcery-ai Bot Apr 23, 2026

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suggestion (bug_risk): Use explicit is None check for buffer to avoid relying on MessageChain truthiness.

Since buffer starts as None, using if not buffer: will also trigger if MessageChain is a valid but falsy value (e.g., empty if it implements __bool__/__len__). That changes behavior between None and an empty chain. Use if buffer is None: to distinguish the initial state from an empty MessageChain and make the intent explicit.

else:
buffer.chain.extend(chain.chain)
if not buffer:
return None
buffer.squash_plain()
await self.send(buffer)
return await super().send_streaming(generator, use_fallback)
Comment on lines +225 to +226
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@sourcery-ai sourcery-ai Bot Apr 23, 2026

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issue (bug_risk): Avoid double-processing and using an already-consumed generator in send_streaming.

Here the generator is fully consumed into buffer, sent once via self.send(buffer), and then passed to super().send_streaming(generator, use_fallback) after it is already exhausted. This makes the superclass call effectively a no-op or fragile if it expects a fresh stream, and risks duplicate or inconsistent output if it also sends content.

Given the docstring that KOOK does not support streaming and this method should fall back to a normal send, consider either:

  • Only aggregating and calling await self.send(buffer) (remove the super().send_streaming call), or
  • Skipping manual aggregation and just return await super().send_streaming(generator, use_fallback) if the base class already provides the fallback.

This avoids using an exhausted generator and prevents double-sending.

Comment on lines +225 to +226
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@gemini-code-assist gemini-code-assist Bot Apr 23, 2026

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high

在此处调用 super().send_streaming(generator, use_fallback) 是不正确的，因为 generator 已经在前面的 async for 循环中被完全耗尽（exhausted）。父类方法将接收到一个空的生成器，这可能导致预期的逻辑（如指标统计或默认处理）失效。由于你已经通过 await self.send(buffer) 发送了聚合后的消息，且 self.send 内部已经调用了 super().send() 来处理通用的发送逻辑（如指标上报），因此这里不需要再调用父类的流式发送方法。

Suggested change
await self.send(buffer)
return await super().send_streaming(generator, use_fallback)
await self.send(buffer)

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